Question

Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0.

Options are:-
1.) (-6,5)
2.) (5,6)
3.) (-5,6)
4.) (6,5)​

Answer 1

Solution :

Let the foot of the perpendicular be M (x¹ , y¹).

• Slope of line AB, i.e., y = -x + 11 = -1

• Slope of line PM = (y¹ - 3)/(x¹ - 2)

Now, PM ⊥ AB

=> (y¹ - 3)/(x¹ - 2) × -1 = -1

=> y¹ - 3 = x¹ - 2

=> x¹ - y¹ = -1 ...i)

Also, (x¹ , y¹) lies on AB.

=> x¹ + y¹ - 11 = 0

=> x¹ + y¹ = 11 ...ii)

Add eq i) and eq ii) :

=> (x¹ - y¹) + (x¹ + y¹) = - 1 + 11

=> x¹ - y¹ + x¹ + y¹ = 10

=> 2x¹ = 10

=> x¹ = 5

By putting x¹ in eq ii) we get :

=> x¹ + y¹ = 11

=> 5 + y¹ = 11

=> y¹ = 11 - 5

=> y¹ = 6

Hence, required foot is (5 , 6).

Answer 2

Given: A perpendicular is drawn from the point (2, 3) on the line x + y - 11 = 0.

To find: We have to find the coordinates of the foot of the perpendicular.

Solution:

Let the coordinates of the foot of the perpendicular is (a, b).

Now the slope of the line AB is-

x+y=11

x=-y+11

The slope is equal to -1.

The slope of line PM is equal to (b-3)/(a-2).

Now given that PM is perpendicular to the AB.

So, we can write (b-3)/(a-2)×-1=-1

b-3=a-2

a-b=-1. (i)

This lies in the AB line so, we can write-

a+b=11. (ii)

By adding the two-expression we get-

a-b+a+b=-1+11

2a=10

a=5

Putting the value of a in expression (i) we get-

5-b=-1

b=6

The coordinates of the foot of the perpendicular are (5, 6).The correct option is 2.